Calculate the electronegativity of fluorine from following data :
`E_(H - H) = 104.2` kcal `mol^(-1)`
`E_(F - F) = 36.6` kcal `mol^(-1)`
`E_(H -F) = 134.6` kcal `mol^(-1)`
Electronegativity of H is 2.05.

Explanation

Energy for 100% covalent bond (H-F)= `sqrt(E_(H-H) xx E_(F-F))`
=`sqrt(104.2 xx 36.6)`
=61.75 kcal
Resonance energy `(Delta)=` acutal bond energy- energy for 100% covalent bond
=134.6-61.75
=72.85 kcal
Electronegativity difference= `0.18 sqrt(Delta)`
Electronegativity of F- electronegativity of `H= 0.18 sqrt(Delta)`
Electronegativity of `F= 0.18 xx sqrt(72.85) + 2.05` = 3.586